Download Haplotype - Biomedical Informatics

BMI 731- Winter 2004
Chapter1: SNP Analysis
Catalin Barbacioru
Department of Biomedical Informatics
Ohio State University
Glossary
• Allele = Alternative form of a gene. One of the different
forms of a gene that can exist at a single locus.
• Genotype = The specific allelic composition of a cell,
either of the entire cell or more commonly for a certain
gene or a set of genes.
• Haplotype = A set of closely linked genetic markers
present on one chromosome which tend to be inherited
together (not easily separable by recombination).
Glossary
• Hardy-Weinberg equilibrium = The stable frequency
distribution of genotypes, AA, Aa, and aa, in the
proportions p^2, 2pq, and q^2 respectively (where p and
q are the frequencies of the alleles, A and a) that is a
consequence of random mating in the absence of
mutation, migration, natural selection, or random drift.
• Linkage disequilibrium = When the observed
frequencies of haplotypes in a population does not agree
with haplotype frequencies predicted by multiplying
together the frequency of individual genetic markers in
each haplotype.
A Little Population Genetics
• Population genetics (and evolutionary genetics) deal with
groups of organisms and families, usually natural
populations.
• We can discern two strands of thought in the area. One is
the study of very large ("ideal") idealized groups or
populations, where models can be deterministic.
• The other is dealing with smaller populations, where the
role of chance can play a larger role (so called genetic
drift).
Genotype and allele frequencies
One question of crucial interest is this: how common are the
different alleles at a given locus in a given population.
The percentages are our best estimate of the probability
that an individual will carry that genotype in the population
of London, Oxford and Cambridge. The observed
heterozygosity is 49.6%.
There is another population described in this table. It is the
population of gametes that gave rise to individuals tested:
The percentages here are our best estimate of the
probability that a sperm or egg taken from that population
will carry that particular allele. If the frequency of the
commonest allele at a particular locus is less than 99%, we
call this a polymorphic locus or polymorphism.
Hardy-Weinberg equilibrium
• Hardy-Weinberg equilibrium describes the relationship
between the gametic or allele frequencies, and the
resulting genotypic frequencies. It holds if the following
properties are true for the given locus,
1.Random mating or panmixia: the choice of a mate is not
influenced by his/her genotype at the locus.
2.The locus does not affect the chance of mating at all,
either by altering fertility or decreasing survival to
reproductive age.
If these properties hold, then the probability that two
gametes will meet and give rise to a new genotype is
simply the product of the allele frequencies (a la
binomial):
P(AA)= P(A) x P(A) = pA2
P(aa)= P(a) x P(a) = pa2
P(Aa)= 1 - P(AA) - P(aa) = 2 x P(A) x P(a)
= 2pApa.
Tests for HWE
For a two-allele case, disequilibrium coefficient is :
D = PAA – pA2
where PAA = P(AA) the probability of AA genotype and
pA = P(A) is the probability of allele A.
If nAA, nAa, naa are the numbers of individuals with
genotypes AA, Aa and aa respectively, from a total of n
individuals, then estimators of the above probabilities are:
PAA = nAA/n, PAa = nAa/n, Paa = naa/n, where n =nAA+nAa+naa
pA = (2nAA+nAa)/2n, pa = (2naa+nAa)/2n and pa + pA = 1
Chi-square test
for HWE
Then under HWE
Genotype
AA
Aa
aa
Observed
nAA
nAa
naa
Expected
npA2
2npApa
npa2
Obs-Exp
nD
-2nD
nD
Chi-square test
for HWE
The goodness-of-fit chi-squared statistic is
XA2 = Σgenotypes (Obs-Exp)2/Exp
= (nD)2/npA2 + (-2nD)2/2npApa + (nD)2/npa2
= nD2/pA2(1-pA)2
and the test rejects (H0) the assumption of HWE if
XA2 > 3.84
The usual problems associated with this test that it is
sensitive to small expected values. An alternative version
(Yates), which overcomes continuity assumptions is:
XA2 = Σgenotypes (|Obs-Exp|-0.5)2/Exp
Fisher (exact) test
for HWE
Under HWE hypothesis, the probability of the observed set
of genotypic counts nAA, nAa and naa in a sample of size n is
P(n AA , n Aa , naa ) 
n!
2
2
( p A ) n AA (2 p A p a ) n Aa ( p a ) naa
n AA !n Aa !naa !
whereas the allele counts nA and na are binomially distributed
if HWE holds:
(2n)!
P(n A , na ) 
( p A ) n A ( p a ) na
n A !na !
Fisher (exact) test
for HWE
Putting together, the probability of the observed genotypic
frequencies, assuming HWE, conditional on the observed
allele frequencies is
P(n AA , n Aa , naa , n A , na )
n!n A !na !2 n Aa
P(n AA , n Aa , naa | n A , na ) 

P(n A , na )
n AA!n Aa !naa !(2n)!
which can be expressed in terms of the allele A number and
Of the number of heterozygotes nAa. We reject the HWE
hypothesis if the above conditional probability is less than
the significance level of type I error (α), usually 0.05.
HWE test - Example
AA
Aa
aa
D
Probability(exact) Chi-square
9
1
30
0.1686
0.0000*
34.67*
8
3
29
0.1436
0.0000*
25.15*
7
5
28
0.1186
0.0001*
17.16*
6
7
27
0.0936
0.0024*
10.68*
5
9
26
0.0686
0.0229*
5.74*
0
19
21
-0.056
0.0823
3.88*
4
11
25
0.043
0.1793
2.32
1
17
22
-0.031
0.4101
1.20
3
13
24
0.018
0.6585
0.42
2
15
23
-0.006
1.0000
0.05
* Causes rejection of HWE at 5% significance level
Power and sample size of tests for HWE
•Statistical tests of hypothesis are subject to two kind of
errors: a true hypothesis may be rejected (type I error or α
or significance level or p-value) or a false hypothesis may
not be rejected (type II error or β or 1-power of the test).
•For the chi-square test, theory provides that, in large
samples, X2 is distributed approximately as a chi-square
with 1 d.f. when the hypothesis is true and as a noncentral
chi-square when the hypothesis is false i.e.
X2 ~ Χ2(1) when H0 is true
X2 ~ Χ2(1, λ) when H0 is false
where λ is the noncentrality parameter (see tables).
Power and sample size of tests for HWE
The disequilibrium coefficient, D, required for attaining 90%
power and a 0.05 significance level for the chi-square test is
D  p A (1  p A )
10.5
n
Alternatively, the number of samples required in order to
attain 90% power and a 0.05 significance level for the
chi-square test when the disequilibrium coefficient is D, is
p A (1  p A ) 2
n  10.5
D2
2
* If the required power is 50% or 80%, then 10.5 is replaced by 3.84 or 8.7
Linkage disequilibrium
Gametic disequilibrium at two loci
Measures the association of two alleles at two different loci.
Given two biallelic loci with alleles A, a and B, b respectively,
let the disequilibrium coefficient be
DAB = pAB – pApB.
The (ML) estimator of DAB is DAB = pAB – pApB.
A chi-square statistic for the hypothesis of no disequilibrium,
H0: DAB=0, is the test statistic
2
X AB
2
2nD AB

p A (1  p A ) p B (1  p B )
and the test rejects H0 if XAB2 > 3.84 .
Linkage disequilibrium
Gametic disequilibrium at two loci
An exact test for gametic linkage disequilibrium depends on
the probabilities of all possible samples of gametic numbers
for the observed allele numbers. Under the assumption of no
linkage disequilibrium
P(n AB )  P(n AB , n Ab , naB, nab )
(2n)!( p A p B ) n AB ( p A pb ) n Ab ( p a p B ) naB ( p a pb ) nab

n AB !n Ab !naB !nab !
and the allele probabilities are
P(n A , na ) 
(2n)!
( p A ) n A ( p a ) na
n A !na !
P(n A , na ) 
(2n)!
( p B ) nB ( pb ) nb
n B ! nb !
Linkage disequilibrium
Gametic disequilibrium at two loci
Taking the ratio between these quantities gives the probability
of gametic numbers conditional on allele numbers:
P(n AB | n A , nB ) 
n A !na !nB !nb !
n AB !n Ab !naB !nab !(2n)!
which depends on n, nAB, nA and nB only. As in the case of
HWE, this probability is compared with the chosen significance
Level (p-value).
Linkage disequilibrium
Genotypic disequilibrium
When genotypes are scored, it is often not possible to
distinguish between the two double heterozygotes AB/ab
and Ab/aB, so that the gametic frequencies cannot be
inferred. Under the assumption of random mating, in which
genotypic frequencies are assumed to be the products of
gametic frequencies, it is possible to estimate gametic
frequencies. A measure of (digenic) linkage disequilibrium
between alleles A and B is:
 AB  2 P
AB
AB
P
AB
Ab
P
AB
aB
1 AB
 ( Pab  PaBAb )  2 p A p B
2
Linkage disequilibrium
Genotypic disequilibrium
If the 9 genotypic classes
are numbered as
BB
Bb
bb
AA
n1
n2
n3
Aa
n4
n5
n6
aa
n7
n8
n9
then an (ML) estimator for ΔAB is:

 AB


1
1
 (2n1  n2  n4  n5 )  2 p A p B
n
2
Linkage disequilibrium
Genotypic disequilibrium
The chi-square test statistics for LD is
_
n  AB

X AB ( A  D A )( B  DB ) ,
2
2
 A  p A (1  p A ),  B  p B (1  p B )
D A  PAA  p A , DB  PBB  p B
2
2
Note the explicit way in which departures from HW are
Included in this expresion.
Δ2 represents the statistical correlation between two sites,
and takes value 1 if only two haplotypes are present. It is
arguably the most relevant measure for association
between susceptibility loci and SNPs. For example,
suppose SNP1 is involved in disease susceptibility, but we
genotype cases and controls at a nearby site SNP2. Then,
to achieve the same power to detect associations at SNP2
as we would have at SNP1, we need to increase our
sample size by a factor of 1/ Δ2.
These measures are defined for pairs of sites, but for some
applications we might instead want to measure how strong
LD is across an entire region that contains many
polymorphic sites — for example, for testing whether the
strength of LD differs significantly among loci or across
populations, or whether there is more or less LD in a region
than predicted under a particular model. Measuring LD
across a region is not straightforward, but one approach is
to use the measure ρ, which measures how much
recombination would be required under a particular
population model to generate the LD that is seen in the
data. The development of methods for estimating is now
an active research. This type of method can potentially also
provide a statistically rigorous approach to the problem of
determining whether LD data provide evidence for the
presence of hotspots.